Let $R$ be a Noetherian ring, and $M$ a finite $R$-module. Then $M$ is torsionless if and only if it is a submodule of a finite free module, where torsionless is defined here. (Bruns and Herzog, Exercise 1.4.20(b).)
Sufficiency is obvious. And I know that $M$ is torsion-free as well. But how to embed $M$ into a finite free module?
By definition, $M$ is torsion-less if the $R$-dual $M^\vee=\mathrm{Hom}_R(M,R)$ separates points, i.e., if the map $M\to\prod_{\varphi\in M^\vee}R$ given by $m\mapsto(\varphi(m))_{\varphi\in M^\vee}$ is injective. But $M^\vee$ is a finitely generated $R$-module. Indeed, we can choose $R^n\to M$ surjective, and then $\mathrm{Hom}_R(M,R)\to\mathrm{Hom}_R(R^n,R)=R^n$ is injective, and since $R$ is Noetherian, submodules of finite $R$-modules are finite, so $M^\vee$ is finitely generated. Let $\varphi_1,\ldots,\varphi_n$ be generators. Then I claim that the map $M\to\bigoplus_{i=1}^n R$ given by $m\mapsto(\varphi_i(m))$ is injective. Suppose that $m$ lies in the kernel, so $\varphi_i(m)=0$ for all $i$. Take any element $\varphi\in M^\vee$ and write $\varphi=\sum_i r_i\varphi_i$. Then $\varphi(m)=\sum_i r_i\varphi(m)=0$. So $M^\vee$ annihilates $m$, and thus $m=0$ by the assumption that $M$ is torsion-less.